Question

A position function \(\vec{r}(t)\) of an object is given. Find the speed of the object in terms of \(t,\) and find where the speed is minimized/maximized on the indicated interval. $$ \vec{r}(t)=\langle 5 \cos t, 5 \sin t\rangle \text { on }[0,2 \pi] $$

Short answer

The speed of the object is 5 and it is constant over the interval [0, 2π].

Step-by-step solution

Find the Velocity Function

The velocity function \(\vec{v}(t)\) is the derivative of the position function \(\vec{r}(t)\) with respect to time \(t\). Calculate \(\vec{v}(t) = \frac{d}{dt}\langle 5 \cos t, 5 \sin t \rangle \). The derivative of \(5 \cos t\) is \(-5 \sin t\) and the derivative of \(5 \sin t\) is \(5 \cos t\). Therefore, \(\vec{v}(t) = \langle -5 \sin t, 5 \cos t \rangle\).

Calculate the Speed Function

The speed of the object is the magnitude of the velocity vector \(\vec{v}(t)\). Compute the magnitude: \[ |\vec{v}(t)| = \sqrt{(-5 \sin t)^2 + (5 \cos t)^2} \]This simplifies to \[ |\vec{v}(t)| = \sqrt{25 \sin^2 t + 25 \cos^2 t} \]Using the Pythagorean identity \(\sin^2 t + \cos^2 t = 1\), we get \[ |\vec{v}(t)| = \sqrt{25} = 5 \]

Determine Where Speed is Maximized/Minimized

Since the speed function \(|\vec{v}(t)| = 5\) is constant for all \(t\), the speed does not vary over the interval \([0, 2\pi]\). Hence, there is no maximum or minimum speed apart from this constant value.

Key concepts

Position Function

In the world of calculus, a position function, denoted as \( \vec{r}(t) \), describes the location of an object in space as it changes over time. This function is often expressed as a vector, with components that represent the coordinates in each dimension. For example, the position function given in the exercise is \( \vec{r}(t) = \langle 5 \cos t, 5 \sin t \rangle \). This describes a circular path with radius 5, as the object moves in a 2D plane. The components, \(5 \cos t\) and \(5 \sin t\), map the x and y positions respectively, tracing out a circle as \(t\) varies.
In practical terms, as \(t\) moves from 0 to \(2\pi \), the object completes one full circle along its path. This is due to the cyclical nature of the trigonometric functions \(\cos\) and \(\sin\), which repeat their values every \(2\pi\). Understanding the position function is key to predicting where an object will be at a given time.

Velocity Vector

The velocity vector, \( \vec{v}(t) \), gives us both the direction and speed of an object's movement. It is essential for understanding how the object's position changes over time. To find this vector, you differentiate the position function with respect to time. In our given exercise, the position function is \( \vec{r}(t) = \langle 5 \cos t, 5 \sin t \rangle \). When differentiating, you apply the derivative rules for trigonometric functions. The derivative of \(5 \cos t\) is \(-5 \sin t\), and the derivative of \(5 \sin t\) is \(5 \cos t\).
Thus, the resulting velocity vector is \( \vec{v}(t) = \langle -5 \sin t, 5 \cos t \rangle \). This vector tells us the object's instantaneous rate of change of position in both the x and y directions. The sign of the components indicates the direction - negative \(-5 \sin t\) suggests movement in the negative x-direction when \(\sin t\) is positive, and positive \(5 \cos t\) suggests movement in the positive y-direction when \(\cos t\) is positive.
Understanding the velocity vector is crucial for assessing how quickly and in what direction the object is moving at any point in time.

Magnitude of Velocity

To find an object's speed, which is a scalar quantity, we calculate the magnitude of the velocity vector. The speed is the length or norm of this vector. It is determined by taking the square root of the sum of the squares of the components of the velocity vector \( \vec{v}(t) \). In this exercise:

• The velocity vector is \( \vec{v}(t) = \langle -5 \sin t, 5 \cos t \rangle \).
• The magnitude is calculated as \(|\vec{v}(t)| = \sqrt{(-5 \sin t)^2 + (5 \cos t)^2}\).

Applying the Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \), this simplifies to \(|\vec{v}(t)| = \sqrt{25}\), which results in a constant speed of 5 for the object's entire motion.
This calculation indicates that the object moves at a constant speed as it follows its circular path; the speed does not change with time. The magnitude of velocity provides a clear measure of the object's pace without direction, crucial for understanding how quickly it progresses along its trajectory.